package programmercarl.动态规划.C48;

class Solution {
    public int numDistinct(String s, String t) {
        //dp含义是以i-1和j-1为结尾的s和t的的不同子序列的个数
        int[][] dp = new int[s.length() + 1][t.length() + 1];
        for (int i = 0; i < s.length(); i++) {
            dp[i][0] = 1;
        }
        for (int i=1; i<=s.length(); i++) {
            for (int j=1; j<=t.length(); j++) {
                if (s.charAt(i-1) == t.charAt(j-1)) {
                    dp[i][j] = dp[i-1][j-1]+dp[i-1][j];
                }else {
                    dp[i][j] = dp[i-1][j];
                }
            }
        }
        return dp[s.length()][t.length()];
    }
}